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I was asked to verify the relation $\partial_t(\nabla \times v)=\nabla \times (v \times (\nabla \times v))$ of an incompressible fluid flow, where $v$ is the flow velocity. I was looking at Landau and Lifshitz "Fluid Mechanics" and they prove this relation holds for isentropic flow. Later, when discussing incompressible flow they assume this identity for incompressible flow without explaining it, only saying

Since the density is no longer an unknown function as it was in the general case, the fundamental system of equations in fluid dynamics for an incompressible fluid can be taken to be equations involving the velocity only. These may be the equation of continuity (10.2) and equation (2.11)

Where equation 2.11 is the desired equation. Should I understand from this that isentropic flow implies incompressibility? Or is there something else at hand?

Qmechanic
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Lourenco Entrudo
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  • Related/useful: https://physics.stackexchange.com/q/729637/226902 https://physics.stackexchange.com/q/745716/226902 https://physics.stackexchange.com/q/507526/226902 – Quillo Jan 16 '23 at 10:57

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No. For a viscous fluid, flow isn't isentropic. Viscosity dissipates kinetic energy as heat, generating entropy.

John Doty
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    Of course for a viscous fluid, equation $\partial_t(\nabla \times v)=\nabla \times (v \times (\nabla \times v))$ does not hold. There is an extra term describing the effect of viscosity. – Alfred May 16 '22 at 00:42